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Answer: 20%
Using inverse operations
100% + 25% = 1 + $\frac14$ = $\frac54$
time $\times$ speed = distance
$\times \frac54$ $\times$ ? $\times$ 1 because distance does not change
(new time) $\times$ (new speed) = distance
$\frac 54 \times$ ? = 1, so ? = $\frac45$ = 80%
80% of original time is a 20% reduction
Using a speed-time graph
The area on a speed-time graph represents distance.
Green rectangle - normal day
Yellow rectangle - traffic jam
The rectangles have the same area since the distance is still the same
Horizontal scale factor $\times$ 125% = 1.25
So vertical scale factor is the inverse
$\div$ 1.25 = $\div \frac54$ = $\times \frac45$ = $\times$ 80%
80% of original time is a 20% reduction
Using algebra
Write $d$ for the distance of Emily's journey, and $t$ for the time it usually takes her. Then, in normal circumstances, her average speed is $\frac{d}{t}$.
Yesterday, her journey took $25\%$ longer than usual, meaning an increase of $0.25t$, so the time was $1.25t$. The distance was the same as usual.
Her average speed was therefore $\frac{d}{1.25t} = \frac{1}{1.25} \times \frac{d}{t} = 0.8\frac{d}{t}$. This means she travelled at $0.8$ of her usual average speed, which is a reduction of $0.2$ or $20\%$.
Traffic Jam
Age 14 to 16
ShortChallenge Level 
Answer: 20%
Using inverse operations
100% + 25% = 1 + $\frac14$ = $\frac54$
time $\times$ speed = distance
$\times \frac54$ $\times$ ? $\times$ 1 because distance does not change(new time) $\times$ (new speed) = distance
$\frac 54 \times$ ? = 1, so ? = $\frac45$ = 80%
80% of original time is a 20% reduction
Using a speed-time graph
The area on a speed-time graph represents distance.
Green rectangle - normal day
Yellow rectangle - traffic jam
The rectangles have the same area since the distance is still the sameHorizontal scale factor $\times$ 125% = 1.25
So vertical scale factor is the inverse
$\div$ 1.25 = $\div \frac54$ = $\times \frac45$ = $\times$ 80%
80% of original time is a 20% reduction
Using algebra
Write $d$ for the distance of Emily's journey, and $t$ for the time it usually takes her. Then, in normal circumstances, her average speed is $\frac{d}{t}$.
Yesterday, her journey took $25\%$ longer than usual, meaning an increase of $0.25t$, so the time was $1.25t$. The distance was the same as usual.
Her average speed was therefore $\frac{d}{1.25t} = \frac{1}{1.25} \times \frac{d}{t} = 0.8\frac{d}{t}$. This means she travelled at $0.8$ of her usual average speed, which is a reduction of $0.2$ or $20\%$.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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